If the determinant of a matrix A, $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ = -3, calculate the determinant of matrix B, $\begin{bmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{bmatrix}$.
I have no idea how to approach this problem.
I can see that the matrix A is contained in B and transposed with a row switch.
$\begin{bmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{bmatrix}^T$ = $\begin{bmatrix}2&c+1&d-2\\-2&-1&2\\0&2a&2b\end{bmatrix}$
Then after some row switches: $R_2 \leftrightarrow R_3$, then $R_1 \leftrightarrow R_3$
$\begin{bmatrix}-2&-1&2\\0&2a&2b\\2&c+1&d-2\end{bmatrix}$
And with row operation $R_3+R_1$..
$\begin{bmatrix}-2&-1&2\\0&2a&2b\\0&c&d\end{bmatrix}$
I can calculate the determinant as $(-2) \cdot (-3) \cdot (2) \cdot (-1)^2$ using cofactor expansion and the properties of determinants. The determinant of the transposed matrix is equal to the determinant of the matrix, the row switches are accounted for by the $(-1)^2$, the row multiplied by 2 is accounted for by $(2)$.
So my answer is 12. Please help me if there's a mistake.
Thank you!
Answer
A determinant is an alternate multilinear function of its columns (and of its rows as well). So
\begin{alignat}{2}
\begin{vmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{vmatrix}&=\begin{vmatrix}0&-2&0\\c &-1&2a\\d &2&2b\end{vmatrix} +\underbrace{\begin{vmatrix}2&-2&0\\1&-1&2a\\-2&2&2b\end{vmatrix}}_{=0 \text{ since columns 1}\\\text{and 2 are collinear}}\\
&=2\begin{vmatrix}0&-2&0\\c &-1&a\\d &2&b\end{vmatrix}=2\cdot 2\begin{vmatrix}c &a\\d&b\end{vmatrix}&\quad&\text{(expanding by the 1st row)}\\&=4(bc-ad).
\end{alignat}
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